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Non relativistic approach

  1. Jun 29, 2015 #1
    1. The problem statement, all variables and given/known data:
    1) Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 x 109 constant, Z = 2 and (hfPlanck's Constant) = 6.6 x 10-34 J s]

    A) 2.92xl06m/s B) 1.46xl06m/s
    C) 0.73xl06m/s D) 3.0xl06m/s



    2. Relevant equations
    V=C/137 multiplied by Z/N
    Here V=velocity of electron
    C=3 multiplied by 10^8
    N=number of orbit
    or 2.18 multiplied by 10^6

    3. The attempt at a solution

    Now if i will solve for this i will get my answer as option B.But I want to know what non-relativistic approach means?I have googled it but did not get anythIng.PLEASE GUIDE..
     
  2. jcsd
  3. Jun 29, 2015 #2

    blue_leaf77

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    I would say, it means just the ordinary (non-relativistic) Schrödinger equation. A more accurate (but not the most accurate existing) theory must include the relativistic effect, this is accounted for in the Dirac equation. But the relativistic effect is very negligible for small Z systems such as He+.
     
  4. Jun 29, 2015 #3
    Sorry but I did not understand.
     
  5. Jun 29, 2015 #4

    blue_leaf77

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    Actually when one says "orbital velocity", it refers to the expectation value of kinetic energy ##\frac{p^2}{2m}## in the state given by the orbital in question. For the non-relativistic case where the Schroedinger equation simply reads as
    $$\left( \frac{p^2}{2m} + V(r)\right)\psi_{nlm} = E_n \psi_{nlm}$$
    it can be shown that ##\langle T \rangle_{nlm} = \langle \frac{p^2}{2m} \rangle_{nlm} = -E_n##. Using this, the orbital velocity is taken as the root-mean-square velocity in that orbital,
    $$ v_{rms} = \sqrt{\langle v^2 \rangle_{nlm}} = \sqrt{ \frac{2 \langle T \rangle_{nlm} }{m} } = \frac{\alpha Z}{n}c $$
    Now if the relativistic were to be taken into account, we must either add correction terms (and subsequently use perturbation method) or directly resort to the so-called Dirac equation, the former is useful when relativistic effect is not too large to be taken as a mere perturbation while the latter corresponds to the strong relativistic effect. But for this problem using perturbation method is more useful. To proceed one must first add the correction terms to the non-relativistic Schroedinger equation, there are more than one of such quantities but only one which directly pertains our problem, namely the kinetic energy correction term
    $$ H_1 = -\frac{p^4}{8m^3c^2}$$
    this correction term can be derived by applying some approximation to the Dirac equation. Therefore, the expectation value of the kinetic energy taking relativistic effect into account should be ##\langle T+H_1 \rangle_{nlm}##. The second term can be calculated analytically, for example in http://quantummechanics.ucsd.edu/ph130a/130_notes/node345.html
    From that link you should see that in the case of weak relativistic effect, the orbital velocity also depends on the orbital quantum number ##l## as opposed to the non-relativistic one.
     
  6. Jun 29, 2015 #5
    This makes more sense now.Thanks @ blue leaf 77 :smile:
     
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